Consider the sublimation of iodine at 25.0 °C : I 2 (s) → I 2 (g) b. Find ΔG° rxn at 25.0 °C under the following nonstandard conditions: i. P I2 = 1.00 mmHg ii. P I2 = 0.100 mmHg

Welcome back everyone to another video, consider the sublimation of iodine at 25 °C. We're given the reaction I two solid forms I two gas B find the change in Gibbs free energy of the reaction at 25 °C under the following non-standard conditions. We're given two cases. The partial pressure of iodine is one millimeter mercury. And then the second case is that the partial pressure of iodine is 0.1 millimeters mercury and therefore answer choices. ABC and D. Those are just variations of the energy values in kilojoules. Now, what we want to do here is just recall the equation that we need to use. If we are looking at non standard conditions, then the Gibbs free energy change delta G is equal to delta G knot, which is the same thing or basically the Gibbs free energy at standard conditions plus RT LNQ. Now we will break everything down just in a bit. But first of all, we understand that to calculate delta G, we need to know delta G not. So we're going to use the constant sables and we will identify that the Gipps free energy at standard conditions. For the sublimation process is equal to 19.3 kilojoules. OK. So that's what we get. This number is obtained from the tables. And now what we're going to do is just understand the reaction quotient Q. First of all, well, Q essentially is similar to the equilibrium constant in terms of its calculation, but the reaction conditions will not necessarily be at equilibrium. So in this case, I will react on this iodine, meaning we're taking the partial pressure of iodine because it is in its gaseous form. And we have its partial pressure and we don't need to divide by the concentrations or pressures of the reactants because we have a solid. So that's our key value. Now T is our temperature. And we have to understand that in this case, if we have 25 °C, this would be 298.15 Kelvin because that's how we convert Celsius to Kelvin, right? We just need to add 273.15 and eventually R is the universal gas constant. So let's solve the first part. Applying this equation, delta G would be equal to delta G, not. So we're taking 19.3 killer jewels. And now we need to take R which is 8.3145 jules per mo per Calvin. Notice how our units are not consistent already. So we can immediately convert joules to kilojoules, taking one kilojoule on top, which is equivalent to 1000 joules. So this is our conversion and then since we have our R value, we need to include temperature. So in this case, our temperature is 298.15 Kelvin. And now we are multiplying this by the natural logarithm of the reaction quotient, which is essentially the partial pressure of iodine for part one, that's one millimeter mercury. But we will convert this N two atmospheres since we want to use atmospheres for our pressure. So let's multiply by a conversion factor, one atmosphere is equivalent to 760 millimeters mercury. And that's it. We have our setup, we will now calculate the result. So now for the first part, we end up with 2.9 kilojoules well done. Now, for the second part, we don't really need to change a lot here because the only thing that changes is the partial pressure of iodine. So what we can do is just use the same thought process here. OK. So we will have the same gives free energy at standard conditions because essentially that's the same process at the same temperature. Now, we have the same R we have the same temperature. It's not changing what we have to modify. In this case is our pressure. So we're going to change that pressure to 0.100 millimeters mercury. And of course, this will affect our answer. We are going to calculate the result in this case. And essentially we end up with negative 2.9 essentially an opposite number to the previous one, right? So we have 2.9 here, negative 2.9 here just because we are decreasing the pressure by a factor of 10 according to the rules of logarithms, that's where it's sensible. And now what we notice is that based on the answer choices, we can identify the correct answer. It would be option D part 12.9 kilojoules, part two, negative 2.9 kilojoules. Thank you for watching.

Ch.19 - Free Energy & Thermodynamics

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Tro 6th Edition

Ch.19 - Free Energy & Thermodynamics

Problem 73b

Chapter 19, Problem 73b

Consider the sublimation of iodine at 25.0 °C : I₂(s) → I₂(g) b. Find ΔG°_rxnat 25.0 °C under the following nonstandard conditions: i. P_I2 = 1.00 mmHg ii. P_I2 = 0.100 mmHg

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Identify the reaction: I_2(s) \rightarrow I_2(g).

Use the equation \( \Delta G = \Delta G^\circ + RT \ln Q \) to find \( \Delta G \) under nonstandard conditions.

Calculate the reaction quotient \( Q \) using \( Q = \frac{P_{I_2}}{P^\circ} \), where \( P^\circ = 1 \text{ atm} \).

For \( P_{I_2} = 1.00 \text{ mmHg} \), convert to atm and calculate \( Q \).

For \( P_{I_2} = 0.100 \text{ mmHg} \), convert to atm and calculate \( Q \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is crucial for predicting the spontaneity of a reaction; a negative ΔG indicates a spontaneous process, while a positive ΔG suggests non-spontaneity. The relationship between ΔG, enthalpy (ΔH), and entropy (ΔS) is given by the equation ΔG = ΔH - TΔS.

Standard State and Nonstandard Conditions

The standard state of a substance is defined as its most stable form at 1 bar (or 1 atm) and a specified temperature, typically 25 °C. Nonstandard conditions refer to any conditions that deviate from these standard states, such as different pressures or concentrations. In the context of the question, the partial pressures of iodine gas (PI2) at 1.00 mmHg and 0.100 mmHg are examples of nonstandard conditions that affect the calculation of ΔG.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant (K) of a reaction to the change in temperature and is often used to derive the Gibbs Free Energy under nonstandard conditions. It can be expressed as ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient. This equation allows for the calculation of ΔG at varying pressures or concentrations, which is essential for solving the given problem regarding the sublimation of iodine.